Why when 127*2 the result is -2 when we change to byte?

Question

This is my code:

public class test {

  public static void main(String[] args) {
    byte a=(byte)127, b=2;    
    byte c=(byte)(a*b);

    System.out.println(c);
  }    
}

Why is the result -2?

Are you aware that your (64-bit?) CPU will have a hard time fiddling around with bytes? — Ingo, Nov 05 '13 at 09:45
See also: [Why does ~True result in -2?](http://stackoverflow.com/questions/21881362) — hippietrail, Feb 20 '14 at 07:41

score 9 · Answer 1 · answered Nov 05 '13 at 09:46

Because a*b will result in a temporary int variable, namely, 254, which is 1111 1110. When cast to byte, this will be handled as a signed value. The MSB is 1, so its value will be negative, and the value will be -(((inv)111 1110) + 1) = -((000 0001) + 1) = -2.

score 3 · Answer 2 · answered Nov 05 '13 at 09:46

a * b is 254, so your code is:

byte c=(byte)254;

The decimal 254 is the binary 11111110, which is: -2. Why?

First of all, the number is negative since it begins with 1 (Two's complement), then:

¬ 1 1 1 1 1 1 1 0 is 0 0 0 0 0 0 0 1.

0 0 0 0 0 0 0 1
              1 +
---------------
0 0 0 0 0 0 1 0

This represents 2 in decimal, but remember that the MSB is 1? So final result is -2.

score 0 · Answer 3 · answered Nov 05 '13 at 09:45

0

Since byte is a signed type 2 * 127 is binary "11111110", which is two's complement for -2.

answered Nov 05 '13 at 09:45

Mithrandir

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score 0 · Answer 4 · answered Nov 05 '13 at 09:45

0

because byte is signed, not unsigned.

254 = FE = 1111 1110

the first '1' represent the number as a negative number.

answered Nov 05 '13 at 09:45

RamonBoza

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score 0 · Answer 5 · answered Nov 05 '13 at 09:51

127 in 8 bits is represented in binary like this:

01111111

Multiply by 2, and you get this:

11111110

(When you multiply by 10 in base 10, all the digits can be shifted left, to the next position. The same is naturally true when you multiply by 2 in binary)

Java uses 2's complement to represent negative numbers. Basically, the left-most bit is the sign bit (0 for +, 1 for -). To convert a positive number in a negative number, flip all the bits and add one.

Example: 00000010 = 2, flip the bits: 11111101, then add one: 11111110 = -2. This is the same as 127*2 above.

score 0 · Answer 6 · answered Nov 05 '13 at 09:58

I think what you tried to ask is why overflow of a 8bits signed integer will turned to negative number for your case.
As in CPU, there is not a 8bits ALU, so 8bits will be expand to 32bits, after operation, return the lower 8bits to user.
In CPU: 0000_007F * 0000_0002 = 0000_00FE
This is a overflow for signed 8bits, but the value has been calculated already.
So returned value is FE.
For 8bits signed, FE is -2.

Why when 127*2 the result is -2 when we change to byte?

6 Answers6

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